# General function PDE

1. Apr 12, 2008

### Mechmathian

1. The problem statement, all variables and given/known data

Does anyone know of how to proove that the solution of the differential equation $$u_{t} = u_{xx}$$ is f(x+t)+ g(x-t) in general functions.

2. Relevant equations

3. The attempt at a solution

It is a pretty easy problem for normal functions, but i have no clue of how to do it in general ones!

2. Apr 12, 2008

### Mechmathian

Sorry, the equation is $$u_{tt} = u_{xx}$$

3. Apr 12, 2008

### quasar987

Notice that with the change of variable a=x+t, b=x-t, the equation becomes u_ab=0, and is then easily solved by integrating twice.

4. Apr 12, 2008

### Coto

This equation is known as the wave equation.. It has several methods of solution including separation of variables, method of characteristics, and method by reduction to canonical form as quasar987 mentioned.

The easiest by far would be quasar's suggestion, and the hardest would be separation of variables since you have to interpret the solution carefully.

By making a change of variables as suggested above, use the chain rule to write u_xx and u_tt in terms of u_aa and u_bb, etc. You end up with u_ab = 0 as quasar has mentioned. Now solve by integration and substitute back in a and b.

Last edited: Apr 12, 2008
5. Apr 12, 2008

### Mechmathian

Thank you for the answers, but I do not think that you take in account the fact that we are looking for solutions in GENERAL functions!! What does it mean to inegrate general functions, what does it mean to substitute variables?? I think that it is much more complex then what you are saying.

6. Apr 12, 2008

### quasar987

Certainly, f(x+t)+ g(x-t) does not solve $$u_{tt} = u_{xx}$$ for EVERY f and g.... For starters, they have to be at least twice differentiable for the statement to even make sense.

So maybe you're misinterpreting the question to some degree.

7. Apr 12, 2008

### Mechmathian

I'm sorry, maybe I am mistaking on the terms.. A general function is a functional on D- the finite, infinitely differentiable functions

8. Apr 12, 2008

### Coto

You'd be surprised.. it actually is exactly as we say. Remember the chain rule? We have a = x + t and b = x - t. So a = a(x,t), b = b(x,t). Now what are u_xx and u_tt. Well in order to do this you need to use the chain rule to write u_xx and u_tt in terms of u_aa, u_bb, u_ab, u_a, and u_b with coefficients which depend on the derivatives of a and b with respect to x and t.

After all said and done you end up with a canonical form u_ab = 0. Integrating w.r.t. a and b gives u = F(a) + G(b). => u = F(x+t) + G(x-t) as required.

9. Apr 12, 2008

### quasar987

Ah, you mean generalized functions.

10. Apr 12, 2008

### Mechmathian

Yeah)) Sorry!!

11. Apr 12, 2008

### quasar987

12. Apr 12, 2008

### Coto

The question seems a bit off. Generalized functions don't seem to naturally pop up here. They do more so for the heat equation.. or PDEs where you encounter kernels more often.

13. Apr 13, 2008

### Mechmathian

What I wrote is definitely true..
The other problem on this theme that was given to us is:

Is it true that a solution of $$u_{t}= u_{x}$$ in generalized functions looks locally like f(t+x)?

14. Apr 13, 2008

### Mechmathian

Does anyone even know a book, where I could read about those knids of problems!?